In this paper we discuss a functional approach to obtain a latticelike structure in
d
algebras, and obtain an exact analog of De Morgan law and some other properties.
1. INTRODUCTION
Y. Imai and K. Iséki introduced two classes of abstract algebras: BCKalgebras and
BCI
algebras (
[8
,
9]
).
BCK
algebras have some connections with other areas: D. Mundici
[13]
proved that
MV
algebras are categorically equivalent to bounded commutative
BCK
algebras, and J. Meng
[11]
proved that implicative commutative semigroups are equivalent to a class of
BCK
algebras. It is well known that bounded commutative
BCK
algebras,
D
posets and
MV
algebras are logically equivalent each other (see [4, p. 420]). We refer useful textbooks for
BCK
/
BCI
algebra to
[4
,
6
,
7
,
12
,
17]
. J. Neggers and H. S. Kim (
[14]
) introduced the notion of dalgebras which is a useful generalization of
BCK
algebras, and then investigated several relations between
d
algebras and
BCK
algebras as well as several other relations between
d
algebras and oriented digraphs. J. S. Han et al. (
[5]
) defined a variety of special
d
algebras, such as strong
d
algebras, (weakly) selective
d
algebras and others. The main assertion is that the squared algebra (
X
; □, 0) of a
d
algebra is a
d
algebra if and only if the root (
X
; ∗, 0) of the squared algebra (
X
; □, 0) is a strong
d
algebra. Recently, the present author with H. S. Kim and J. Neggers (
[10]
) explored properties of the set of
d
units of a
d
algebra. It was noted that many
d
algebras are weakly associative, and the existence of nonweakly associative
d
/
BCK
algebras was demonstrated. Moreover, they discussed the notions of a
d
integral domain and a leftinjectivity in
d
/
BCK
algebras. We refer to
[1
,
2
,
15
,
16]
for more information on
d
algebras.
In this paper we discuss a functional approach to obtain a latticelike structure in
d
algebras, and obtain an exact analog of De Morgan law and some other properties.
2. PRELIMINARIES
An (
ordinary
)
d

algebra
(
[14]
) is a nonempty set
X
with a constant 0 and a binary operation “ ∗ ” satisfying the following axioms:

(D1)x∗x= 0,

(D2) 0 ∗x= 0,

(D3)x∗y= 0 andy∗x= 0 implyx=yfor allx,y∈X.
A
BCK
algebra is a
d
algebra
X
satisfying the following additional axioms:

(D4) (x∗y) ∗ (x∗z)) ∗ (z∗y) = 0,

(D5) (x∗ (x∗y)) ∗y= 0 for allx,y,z∈X.
Example 2.1
(
[14]
). Consider the real numbers
R
, and suppose that (
R
; ∗,
e
) has the multiplication
x
∗
y
= (
x
−
y
)(
x
−
e
) +
e
Then
x
∗
x
=
e
;
e
∗
x
=
e
;
x
∗
y
=
y
∗
x
=
e
yields (
x
−
y
)(
x
−
e
) = 0, (
y
−
x
)(
y
−
e
) =
e
and
x
=
y
or
x
=
e
=
y
, i.e.,
x
=
y
, i.e., (
R
; ∗,
e
) is a
d
algebra.
3. A FUNCTIONAL APPROACH TOdALGEBRAS
Let (
X
, ∗, 0) be a
d
algebra. A map
ϕ
:
X
→
X
is said to be
order reversing
if
x
∗
y
= 0 then
ϕ
(
y
) ∗
ϕ
(
x
) = 0 for all
x
,
y
∈
X
;
selfinverse
if
ϕ
(
ϕ
(
x
)) =
x
for all
x
∈
X
; an
antihomomorphism
if
ϕ
(
x
∗
y
) =
ϕ
(
y
) ∗
ϕ
(
x
) = 0 for all
x
,
y
∈
X
; a
homomorphims
if
ϕ
(
x
∗
y
) =
ϕ
(
x
) ∗
ϕ
(
y
) for all
x
,
y
∈
X
.
Example 3.1.
Consider
X
:= {0,
a
, 1} with
Then (
X
; ∗, 0) is a
d
algebra. If we define a map
ϕ
:
X
→
X
by
ϕ
(0) = 1,
ϕ
(
a
) =
a
and
ϕ
(1) = 0, then it is easy to see that
ϕ
is both selfinverse and order reversing, but it is not an antihomomorphism, since
ϕ
(
a
∗1) =
ϕ
(0) = 1 and
ϕ
(1)∗
ϕ
(
a
) = 0∗
a
= 0.
Moreover, it is not a homomorphism, since
ϕ
(0 ∗
a
) =
ϕ
(0) = 1 ≠
a
= 1 ∗
a
=
ϕ
(0) ∗
ϕ
(
a
).
Proposition 3.2.
Let
(
X
, ∗, 0)
be a dalgebra
.
If ϕ
:
X
→
X
is a (anti) homomorphism, then ϕ
(0) = 0.
Proof
. Since
X
is a
d
algebra, by (
D
1), we obtain
ϕ
(0) =
ϕ
(
x
∗
x
) =
ϕ
(
x
) ∗
ϕ
(
x
) = 0.
Proposition 3.3.
If
(
X
, ∗, 0) i
s a dalgebra, then every anti homomorphism is order reversing
.
Proof
. Let
ϕ
:
X
→
X
be an antihomomorphism. If we assume that
x
∗
y
= 0, then
ϕ
(
y
) ∗
ϕ
(
x
) =
ϕ
(
x
∗
y
) =
ϕ
(0) = 0 by Proposition 3.2. This proves the proposition.
Remark.
The converse of Proposition 3.3 need not be true in general. In Example 3.1, the mapping
ϕ
is an order reversing, but not an antihomomorphism.
Let (
X
, ∗, 0) be a
d
algebra and let
ϕ
:
X
→
X
be a map. We denote by 1 :=
ϕ
(0).
Proposition 3.4.
Let
(
X
, ∗, 0)
be a dalgebra and let ϕ
:
X
→
X be both order reversing and selfinverse
.
Then
(
X
, ∗, 0)
is bounded.
Proof
. Given
x
∈
X
, we have

x∗ 1 =x∗ϕ(0) [1 =ϕ(0)]

=ϕ(ϕ(x)) ∗ϕ(0) [ϕ: selfinverse]

= 0 [ϕ: order reversing]
Let (
X
, ∗, 0) be a
d
algebra. We define a relation “≤” on
X
by
x
≤
y
if and only if
x
∗
y
= 0 for all
x
,
y
∈
X
. Note that the relation ≤ need not be a partial order on
X
. We define a relation “∧ on
X
by
x
∧
y
:=
x
∗ (
x
∗
y
)) for all
x
,
y
∈
X
.
Proposition 3.5.
Let
(
X
, ∗, 0)
be a dalgebra
.
If ϕ
:
X
→
X is selfinverse
,
then ϕ
(1) = 0.
Proof
. It follows from
ϕ
is selfinverse that 0 =
ϕ
(
ϕ
(0)) =
ϕ
(1).
Theorem 3.6.
Let
(
X
, ∗, 0)
be a dalgebra and let ϕ
:
X
→
X be a selfinverse map
.
If we define x
∨
y
:=
ϕ
[
ϕ
(
y
) ∧
ϕ
(
x
)],
then
ϕ
(
x
∧
y
) =
ϕ
(
y
) ∨
ϕ
(
x
)
for all x
,
y
∈
X
.
Proof
. Given
x
,
y
∈
X
, we have

ϕ(x∧y) =ϕ[ϕ(ϕ(x)) ∧ϕ(ϕ(y))] [ϕ: selfinverse]

=ϕ[ϕ(a)) ∧ϕ(b)] [a=ϕ(x),b=ϕ(y)]

=b∨a

=ϕ(y) ∨ϕ(x)
Theorem 3.6 shows that the first De Morgan’s law implies the analog of the second De Morgan’s law and conversely, since
x
∨
y
≠
y
∨
x
in general. Moreover, it follows that
x
∧
y
=
ϕ
(
ϕ
(
x
∧
y
)) =
ϕ
[
ϕ
(
y
) ∨
ϕ
(
x
)] for all
x
,
y
∈
X
.
Theorem 3.7.
Let
(
X
, ∗, 0)
be a dalgebra with
for all x
∈
X
.
If ϕ
:
X
→
X is a selfinverse map
,
then x
∨
x
=
x and x
∧
x
=
x for all x
∈
X
.
Proof
. (i). Given
x
∈
X
, we have

x∨x=ϕ[ϕ(x) ∧ϕ(x)] [Theorem 3.6]

=ϕ[ϕ(x) ∗ (ϕ(x) ∗ϕ(x)]

=ϕ(ϕ(x) ∗ 0) [(D1)]

=ϕ(ϕ(x)) [(1)]

=x[ϕ: selfinverse]
(ii).
x
∧
x
=
x
∗ (
x
∗
x
) =
x
∗ 0 =
x
.
Proposition 3.8.
Let
(
X
, ∗, 0)
be a dalgebra with
for all x
,
y
,
z
∈
X
.
Then x
∧
y
≤
x and x
∧
y
≤
y for all x
,
y
∈
X
.
Proof
. (i). Given
x
,
y
∈
X
, by applying (2), we obtain

(x∧y) ∗a= (x∗ (x∗y)) ∗a

= (x∗x) ∗ (x∗y)

= 0 ∗ (x∗y)

= 0
(ii). Given
x
,
y
∈
X
, we have (
x
∧
y
)∗
y
= (
x
∗(
x
∗
y
))∗
y
= (
x
∗
y
)∗(
x
∗
y
) = 0.
Theorem 3.9.
Let
(
X
, ∗, 0)
be a dalgebra with the condition
(
2
).
If ϕ
:
X
→
X is a selfinverse antihomomorphism
,
then x
∗ (
x
∨
y
) = 1
and y
∗ (
x
∨
y
) = 1 f
or all x
,
y
∈
X
.
Proof
. (i). Since
ϕ
:
X
→
X
is a selfinverse antihomomorphism, for all
x
,
y
∈
X
, we obtain

x∗ (x∨y) =x∗ϕ(ϕ(y) ∧ϕ(x))

=x∗ϕ[ϕ(y) ∗ (ϕ(y) ∗ϕ(x))]

=ϕ(ϕ(x)) ∗ϕ[ϕ(y) ∗ (ϕ(y) ∗ϕ(x))]

=ϕ[[ϕ(y) ∗ (ϕ(y) ∗ϕ(x))] ∗ϕ(x)]

=ϕ[[(ϕ(y) ∗ϕ(x)) ∗ (ϕ(y) ∗ϕ(x))]]

=ϕ(0)

= 1
and

y∗ (x∨y) =ϕ(ϕ(x)) ∗ϕ[ϕ(y) ∗ (ϕ(y) ∗ϕ(x))]

=ϕ[[ϕ(y) ∗ (ϕ(y) ∗ϕ(x))] ∗ϕ(y)]

=ϕ[(ϕ(y) ∗ϕ(y)) ∗ (ϕ(y) ∗ϕ(x))]

=ϕ(0)

= 1
CONCLUSION
Whether such functions exists or not depends on the special properties of the
d
algebras.
BCK
algebras have the partial order structure, but
d
algebras have no such a structure and so we need to seek another conditions for obtaining the analog of structures in
d
algebras. This kind of functional approach can be connected with mirror
d
algebras discussed in
[3]
in a new direction.
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